The Stacks project

Proof. Since every Brauer class is represented by a finite central skew field over $k$, we see that the second statement follows from the first by Lemma 11.8.3.

To prove the first statement, suppose that we are given a separable subfield $k' \subset K$. Then the centralizer $K'$ of $k'$ in $K$ has center $k'$, and the problem reduces to finding a maximal subfield of $K'$ separable over $k'$. Thus it suffices to prove, if $k \not= K$, that we can find an element $x \in K$, $x \not\in k$ which is separable over $k$. This statement is clear in characteristic zero. Hence we may assume that $k$ has characteristic $p > 0$. If the ground field $k$ is finite then, the result is clear as well (because extensions of finite fields are always separable). Thus we may assume that $k$ is an infinite field of positive characteristic.

To get a contradiction assume no element of $K$ is separable over $k$. By the discussion in Fields, Section 9.28 this means the minimal polynomial of any $x \in K$ is of the form $T^ q - a$ where $q$ is a power of $p$ and $a \in k$. Since it is clear that every element of $K$ has a minimal polynomial of degree $\leq \dim _ k(K)$ we conclude that there exists a fixed $p$-power $q$ such that $x^ q \in k$ for all $x \in K$.

Consider the map

and write it out in terms of a $k$-basis $\{ a_1, \ldots , a_ n\} $ of $K$ with $a_1 = 1$. So

Since multiplication on $K$ is $k$-bilinear we see that each $f_ i$ is a polynomial in $x_1, \ldots , x_ n$ (details omitted). The choice of $q$ above and the fact that $k$ is infinite shows that $f_ i$ is identically zero for $i \geq 2$. Hence we see that it remains zero on extending $k$ to its algebraic closure $\overline{k}$. But the algebra $K \otimes _ k \overline{k}$ is a matrix algebra (for example by Lemmas 11.4.9 and 11.5.3), which implies there are some elements whose $q$th power is not central (e.g., $e_{11}$). This is the desired contradiction. $\square$